Problem: What is the value of the following logarithm? $\log_{7} \left(\dfrac{1}{7}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $7^{y} = \dfrac{1}{7}$ Any number raised to the power $-1$ is its reciprocal, so $7^{-1} = \dfrac{1}{7}$ and thus $\log_{7} \left(\dfrac{1}{7}\right) = -1$.